If `sec(theta-alpha), sectheta` and `sec(theta+alpha)` are in AP, where `cos alpha ne 1` then what is the value of `sin^2 theta + cos alpha`?

Options

A. 0
B. 1
C. `-1`
D. `1//2`

Answer

To solve the problem, we need to show that if \( \sec(\theta - \alpha), \sec(\theta), \sec(\theta + \alpha) \) are in Arithmetic Progression (AP), then we can find the value of \( \sin^2 \theta + \cos \alpha \).

### Step-by-Step Solution:

1. **Understanding the condition for AP**:
If three numbers \( a, b, c \) are in AP, then the condition is:
\[ 2b = a + c \]
Here, let \( a = \sec(\theta - \alpha) \), \( b = \sec(\theta) \), and \( c = \sec(\theta + \alpha) \). Therefore, we have:
\[ 2 \sec(\theta) = \sec(\theta - \alpha) + \sec(\theta + \alpha) \]

2. **Expressing secants in terms of cosines**:
Recall that \( \sec(x) = \frac{1}{\cos(x)} \). Thus, we can rewrite the equation:
\[ 2 \cdot \frac{1}{\cos(\theta)} = \frac{1}{\cos(\theta - \alpha)} + \frac{1}{\cos(\theta + \alpha)} \]

3. **Finding a common denominator**:
The common denominator for the right-hand side is \( \cos(\theta - \alpha) \cos(\theta + \alpha) \). Therefore, we can rewrite the equation as:
\[ 2 \cdot \frac{1}{\cos(\theta)} = \frac{\cos(\theta + \alpha) + \cos(\theta - \alpha)}{\cos(\theta - \alpha) \cos(\theta + \alpha)} \]

4. **Using the cosine addition formula**:
We know that:
\[ \cos(\theta + \alpha) + \cos(\theta - \alpha) = 2 \cos(\theta) \cos(\alpha) \]
Substituting this into our equation gives:
\[ 2 \cdot \frac{1}{\cos(\theta)} = \frac{2 \cos(\theta) \cos(\alpha)}{\cos(\theta - \alpha) \cos(\theta + \alpha)} \]

5. **Cross-multiplying**:
Cross-multiplying leads to:
\[ 2 \cos(\theta - \alpha) \cos(\theta + \alpha) = 2 \cos^2(\theta) \cos(\alpha) \]
Dividing both sides by 2:
\[ \cos(\theta - \alpha) \cos(\theta + \alpha) = \cos^2(\theta) \cos(\alpha) \]

6. **Using the cosine product-to-sum identity**:
We can use the identity:
\[ \cos(A) \cos(B) = \frac{1}{2} [\cos(A + B) + \cos(A - B)] \]
Applying this gives:
\[ \frac{1}{2} [\cos(2\theta) + \cos(0)] = \cos^2(\theta) \cos(\alpha) \]
Simplifying, we get:
\[ \frac{1}{2} [\cos(2\theta) + 1] = \cos^2(\theta) \cos(\alpha) \]

7. **Rearranging the equation**:
Rearranging gives:
\[ \cos(2\theta) + 1 = 2 \cos^2(\theta) \cos(\alpha) \]
Since \( \cos(2\theta) = 2\cos^2(\theta) - 1 \), we can substitute:
\[ 2\cos^2(\theta) - 1 + 1 = 2 \cos^2(\theta) \cos(\alpha) \]
This simplifies to:
\[ 2\cos^2(\theta) = 2 \cos^2(\theta) \cos(\alpha) \]

8. **Dividing by \( 2\cos^2(\theta) \)** (assuming \( \cos^2(\theta) \neq 0 \)):
\[ 1 = \cos(\alpha) \]
Since \( \cos(\alpha) \neq 1 \) is given in the problem, we need to find the value of \( \sin^2 \theta + \cos \alpha \).

9. **Using the Pythagorean identity**:
Since \( \cos^2(\theta) + \sin^2(\theta) = 1 \), we can express:
\[ \sin^2(\theta) = 1 - \cos^2(\theta) \]

10. **Final substitution**:
From the previous steps, we found that \( \cos^2(\theta) = 1 + \cos(\alpha) \). Thus:
\[ \sin^2(\theta) + \cos(\alpha) = (1 - (1 + \cos(\alpha))) + \cos(\alpha) = 0 \]

### Conclusion:
The value of \( \sin^2 \theta + \cos \alpha \) is \( 0 \).